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Tambaya 2 Rahoto
Which of the following factors will speed up the rate of evolution of carbon (iv) oxide in the reaction below?
2HCl + CaCO3 → CaCl2 + H2 O + CO2
Bayanin Amsa
The following factors increase a reaction rate
- Increase in concentration of reactants
- Increase in temperature
- Addition of catalyst
- Increase in the surface area of reactant(s)
Tambaya 3 Rahoto
A solution X, on mixing with AgNO3 solution gives a white precipitate soluble in aqueous NH3 , a solution Y, when also added to X, also gives a white precipitate which is soluble when heated solutions X and Y respectively contain
Tambaya 4 Rahoto
Methane is prepared in the laboratory by heating a mixture of sodium ethanoate with soda lime. The chemical constituent(s) of soda lime is/are
Bayanin Amsa
The chemical constituent of soda lime used to prepare methane in the laboratory is Ca(OH)2 (calcium hydroxide) and NaOH (sodium hydroxide). Soda lime is a mixture of these two compounds. When sodium ethanoate (NaC2H3O2) is heated with soda lime, it undergoes a reaction known as the Kolbe's reaction, which produces methane gas (CH4) as one of the products. The reaction can be represented as follows: 2NaC2H3O2 + 2Ca(OH)2 → 2CH4 + 2NaOH + 2CaCO3 In this reaction, the sodium ethanoate reacts with the calcium hydroxide to form calcium acetate (Ca(C2H3O2)2) and sodium hydroxide. The calcium acetate then decomposes to produce methane gas and calcium carbonate (CaCO3), which is a solid precipitate. Therefore, the chemical constituents of soda lime used to prepare methane in the laboratory are calcium hydroxide (Ca(OH)2) and sodium hydroxide (NaOH).
Tambaya 5 Rahoto
Which of the following statements does not show Rutherford's account of Nuclear Theory? An atom contains a region
Bayanin Amsa
Rutherford's account of Nuclear theory does not include the fact that atoms contain a massive region and cause deflection of from projectiles.
Tambaya 7 Rahoto
The two ions responsible for hardness in water are
Bayanin Amsa
The ions responsible for hardness in water are Ca2+ and/or Mg2+. Hardness in water refers to the presence of calcium and magnesium ions, which are commonly found in natural water sources such as rivers, lakes, and groundwater. These ions can react with soap to form insoluble compounds, reducing the effectiveness of soap and causing scaling in pipes and appliances. The hardness of water is often measured in terms of the concentration of calcium and magnesium ions, expressed as calcium carbonate equivalents (CaCO3).
Tambaya 8 Rahoto
The IUPAC name for CH3 CH2 COOCH2 CH3 is
Bayanin Amsa
The IUPAC name for the given molecule is ethyl propanoate. To arrive at the IUPAC name, we first identify the longest continuous chain of carbon atoms, which in this case is a 4-carbon chain (propane). We then identify and name the substituent groups attached to this chain, which are a methyl group (CH3) attached to the second carbon atom and an ethoxy group (OC2H5) attached to the third carbon atom. The ethoxy group is named as an ethyl group, and the entire molecule is named as ethyl propanoate, following the standard IUPAC naming conventions for esters.
Tambaya 9 Rahoto
Which of the following statements about catalyst is false?
Bayanin Amsa
The false statement about catalysts is: "catalysts do not alter the mechanism of the reaction and never appear in the rate law." Catalysts are substances that speed up chemical reactions without being consumed in the process. They achieve this by reducing the activation energy needed for the reaction to occur. Enzymes are a type of biological catalysts. In a chemical reaction, a catalyst is not consumed and does not appear in the overall balanced equation. However, catalysts can alter the mechanism of a reaction by providing an alternative pathway with a lower activation energy. This alternative pathway can have a different rate-determining step, which means that the presence of the catalyst can change the rate law of the reaction. Therefore, the statement that catalysts do not alter the mechanism of the reaction and never appear in the rate law is false.
Tambaya 10 Rahoto
When ammonia and hydrogen ion bond together to form ammonium ion, the bond formed is called
Bayanin Amsa
When ammonia and hydrogen ion go into bonding, they form ammonium ion by combining with a dative/coordinate covalent bond.
Tambaya 11 Rahoto
The IUPAC nomenclature of the compound
H3 C - CH(CH3 ) - CH(CH3 ) - CH2 - CH3
Tambaya 12 Rahoto
In the reaction between sodium hydroxide and tetraoxosulphate (VI) solutions, what volume of 0.5 molar sodium hydroxide would exactly neutralize 10cm3 of 1.25 molar tetraoxosulphate (vi) acid?
Bayanin Amsa
Equation of reaction : 2NaOH + H2 SO4 → Na2 SO4 + 2H2 O
Concentration of a base, CB = 0.5M
Volume of acid, VA = 10cm3
Concentration of an acid, CA = 1.25M
Volume of base, VB = ?
Recall:
CAVACBVB=nAnB
... (1)
N.B: From the equation,
nAnB=12
From (1)
1.25×100.5×VB=12
12.50.5VB=12
25 = 0.5VB
VB = 50.0 cm3
Tambaya 13 Rahoto
2-methylprop-1-ene is an isomer of
Bayanin Amsa
2-methylprop-1-ene is an isomer of 3-methyl but-1-ene and 2-methyl but-1-ene. An isomer is a molecule that has the same molecular formula as another molecule, but a different arrangement of atoms. In this case, 2-methylprop-1-ene has the molecular formula C4H8, and so do 3-methyl but-1-ene and 2-methyl but-1-ene. The difference between these three molecules is in the arrangement of the carbon and hydrogen atoms. 2-methylprop-1-ene has a branched structure with a double bond between the first and second carbon atoms. 3-methyl but-1-ene is also a branched molecule, but the double bond is between the second and third carbon atoms. Similarly, 2-methyl but-1-ene has a double bond between the first and second carbon atoms, but it has a different branching pattern. On the other hand, pent-2-ene has five carbon atoms, so it has a different molecular formula than 2-methylprop-1-ene. Therefore, 2-methylprop-1-ene is an isomer of 3-methyl but-1-ene and 2-methyl but-1-ene, but not of pent-2-ene, because it has the same molecular formula and a different arrangement of atoms compared to the other two isomers.
Tambaya 14 Rahoto
Which of the following could not be alkane?
Bayanin Amsa
An alkane is a type of hydrocarbon with only single bonds between the carbon atoms. It follows the general formula CnH2n+2, where "n" is the number of carbon atoms in the molecule. To determine whether a molecule is an alkane or not, we can calculate its molecular formula and check if it fits the general formula of alkane. Out of the given options, the third one (C7H14) cannot be an alkane. To see why, let's use the general formula of alkane, which is CnH2n+2. For C7H14 to be an alkane, it should have 2n+2 = 2(7) + 2 = 16 hydrogen atoms. However, C7H14 has only 14 hydrogen atoms, which means it does not follow the general formula of alkane. Therefore, C7H14 cannot be an alkane. The other options are as follows: - C4H10: This is butane, which is an alkane with four carbon atoms. - C5H12: This is pentane, which is an alkane with five carbon atoms. - C8H18: This is octane, which is an alkane with eight carbon atoms. In summary, the molecule C7H14 cannot be an alkane because it does not follow the general formula of alkane, while the other options are all examples of alkanes.
Tambaya 15 Rahoto
The emission of two successive beta particles from the nucleus 3215P will produce
Bayanin Amsa
Tambaya 16 Rahoto
A cell shorthand notation can be written as A / A+ // B2+ /B. The double slash in the notation represents the
Bayanin Amsa
The double slash in the cell shorthand notation represents the salt bridge. A salt bridge is a component of an electrochemical cell that connects the two half-cells and allows the flow of ions between them. It consists of an inert electrolyte solution (usually a salt) that is placed between the two half-cells. The purpose of the salt bridge is to maintain electrical neutrality in each half-cell by allowing the flow of ions to balance the charge buildup in the half-cells. In the cell shorthand notation, the double slash "//" represents the salt bridge that connects the two half-cells of the electrochemical cell. The first half-cell is represented on the left-hand side of the slash and the second half-cell is represented on the right-hand side of the slash. The anode (where oxidation occurs) is represented on the left side, and the cathode (where reduction occurs) is represented on the right side. Therefore, the correct answer is option number 3: salt bridge.
Tambaya 17 Rahoto
The cost of discharging 6.0g of a divalent metal, X from its salt is ₦12.00. What is the cost of discharging 9.0g of a trivalent metal, Y from its salt under the same condition?
[X = 63, Y = 27, 1F = 96,500C]
Bayanin Amsa
For X: X2+
+ 2e−
→
X
2F = 63g
xF = 6g
x = 6×263=421F
421
F = N12.00
1F = 12421
= N63.00
1F is equivalent to N63.00.
For Y: Y3+
+ 3e−
→
Y
3F = 27g
xF = 9g
x = 3×927
= 1F
1F = N63.00
Tambaya 18 Rahoto
The following are isoelectronic ions except
Bayanin Amsa
Two or more ions are said to be isoelectronic if they have the same electronic structure and the same number of valence electrons.
Na+
= 10 electrons = 2, 8
Mg2+
= 10 electrons = 2,8
O2−
= 10 electrons = 2,8
Si2+
= 12 electrons = 2,8,2
⟹
Si2+
is not isoelectronic with the rest.
Tambaya 19 Rahoto
What mass of magnesium would be obtained by passing a current of 2 amperes for 2 hours, through molten magnesium chloride?
[1 faraday = 96500C, Mg = 24]
Bayanin Amsa
Current (I) = 2A; Time (t) = 2 hours = 7200 secs
Q = It
= 2 x 7200 = 14400C
1 F = 96500C
x = 14400C
x = 1440096500
= 0.15F
Mg2+
+ 2e−
→
Mg
2F →
24g
0.15F →
x
2x = 24 x 0.15
x = 24×0.152
= 1.8g
Tambaya 21 Rahoto
Consider the reaction: A + 2B(g)⇌ 2C + D(g) (Δ H = +ve)
What will be the effect of decrease in temperature on the reaction?
Bayanin Amsa
The effect of a decrease in temperature on the reaction will be that the rate of the backward reaction will increase. In a chemical reaction, the rate of the forward and backward reactions are determined by the activation energy required for each step and the temperature of the system. When the temperature is decreased, the rate of the reaction decreases, and the rate of the backward reaction increases. This shift in the rate of the backward reaction means that there will be a shift in the position of the equilibrium of the reaction. As the rate of the backward reaction increases, the concentration of the reactants will increase and the concentration of the products will decrease, leading to a decrease in the overall yield of the products. In this reaction, as ΔH (the change in enthalpy) is positive, which means that the reaction is endothermic. Endothermic reactions absorb heat from the surroundings to proceed, so a decrease in temperature will lead to a decrease in the rate of the forward reaction and an increase in the rate of the backward reaction. This shift in the rate of the backward reaction will shift the position of the equilibrium of the reaction to the left, leading to an increase in the concentration of the reactants and a decrease in the concentration of the products.
Tambaya 22 Rahoto
Consider the equation below:
Cr2 O2−7 + 6Fe2+ + 14H+ → 2Cr3+ + 6Fe3+ + 7H2 O.
The oxidation number of chromium changes from
Bayanin Amsa
Cr2
O2−7
+ 6Fe2+
+ 14H+
→
2Cr3+
+ 6Fe3+
+ 7H2
O
The oxidation of Cr in Cr2
O2−7
:
Let the oxidation of Cr = x;
2x + (-2 x 7) = -2 ⟹
2x - 14 = -2
2x = 12 ; x = +6
Hence, the change in oxidation of Cr = +6 to +3
Tambaya 23 Rahoto
At 27°C, 58.5g of sodium chloride is present in 250cm3 of a solution. The solubility of sodium chloride at this temperature is?
(molar mass of sodium chloride = 111.0gmol−1 )
Bayanin Amsa
Given the Mass of the salt = 58.5g
Volume = 250 cm3
= 0.25 dm3
Mass concentration = MassVolume
= 58.50.25
= 234 gdm−3
Solubility (in moldm−3
= 234111
= 2.11 moldm−3
≊
2.0 moldm−3
Tambaya 24 Rahoto
Which of the following alkaline metals react more quickly spontaneously with water?
Bayanin Amsa
The alkaline earth metals (Be, Mg, Ca, Sr, Ba, and Ra) are the second most reactive metals in the periodic table, and, like the Group 1 metals, have increasing reactivity in the higher periods. Beryllium (Be) is the only alkaline earth metal that does not react with water or steam, even if metal is heated to red heat. Additionally, beryllium has a resistant outer oxide layer that lowers its reactivity at lower temperatures.
Magnesium shows insignificant reaction with water, but burns vigorously with steam or water vapor to produce white magnesium oxide and hydrogen gas:
A metal reacting with cold water will produce metal hydroxide. However, if a metal reacts with steam, like magnesium, metal oxide is produced as a result of metal hydroxides splitting upon heating.
The hydroxides of calcium, strontium and barium are only slightly water-soluble but produce sufficient hydroxide ions to make the environment basic, giving a general equation of:
| Order of reactivity | Metal | Reactions with water or steam |
|---|---|---|
| most reactive | potassium (K) | very vigorous reaction with cold water |
| ↑ | sodium (Na) | vigorous reaction with cold water |
| ↓ | calcium (Ca) | less vigorous reaction with cold water |
| least reactive | magnesium (Mg) | slow reaction with cold water, vigorous with steam |
Tambaya 25 Rahoto
Which important nitrogen-containing compound is produced in Haber's process?
Bayanin Amsa
The important nitrogen-containing compound that is produced in Haber's process is NH3, which is also known as ammonia. Haber's process is a chemical process used to produce ammonia by reacting nitrogen gas (N2) and hydrogen gas (H2) under high pressure and temperature in the presence of an iron catalyst. The reaction between nitrogen and hydrogen produces ammonia as the main product, along with some nitrogen and hydrogen gases that do not react. NH3 is an important compound that is widely used in industry for the production of fertilizers, plastics, and other chemical products. It is also used as a cleaning agent, a refrigerant, and a fuel for engines. In addition, NH3 is an essential compound for life, as it is a key component of amino acids, which are the building blocks of proteins.
Tambaya 26 Rahoto
A synthetic rubber is obtained from the polymerization of
Bayanin Amsa
A synthetic rubber is obtained from the polymerization of isoprene. Isoprene is a type of hydrocarbon that can be polymerized, or chemically joined together, to form long chains. This process is called polymerization, and the resulting material is called a polymer. When isoprene is polymerized, it forms a synthetic rubber, which is a type of polymer that is used in a wide range of products, including tires, hoses, and adhesives. Synthetic rubber offers several advantages over natural rubber, including improved durability and resistance to heat, ozone, and chemicals.
Tambaya 27 Rahoto
Which quantum divides shells into orbitals?
Bayanin Amsa
The quantum that divides shells into orbitals is the "Azimuthal" quantum number, also known as the "angular momentum" quantum number. The azimuthal quantum number determines the shape of an electron's orbital, which is a region in space where there is a high probability of finding an electron. It describes the angular momentum of an electron in an atom and the number of subshells within a given shell. Each subshell is associated with a specific shape, and can hold a certain number of electrons. The azimuthal quantum number is represented by the letter "l" and can have integer values ranging from 0 to (n-1), where "n" is the principal quantum number. Each value of "l" corresponds to a different subshell shape: - l = 0 corresponds to an "s" subshell, which is spherical in shape. - l = 1 corresponds to a "p" subshell, which has a dumbbell shape with two lobes. - l = 2 corresponds to a "d" subshell, which has a more complex shape with four lobes and a doughnut-like ring. - l = 3 corresponds to an "f" subshell, which has an even more complex shape with eight lobes. The number of orbitals within a subshell is equal to 2l+1. For example, a "p" subshell (l = 1) has three orbitals (2l+1 = 3), which are labeled as "px", "py", and "pz". In summary, the azimuthal quantum number determines the shape of the electron's orbital and the number of subshells within a given shell, and it is represented by the letter "l".
Tambaya 28 Rahoto
When the end alkyl groups of ethyl ethanoate are interchanged, the compound formed is
Bayanin Amsa
The compound formed when the end alkyl groups of ethyl ethanoate are interchanged is ethyl propanoate. This is because ethyl ethanoate consists of two parts: the "ethyl" group and the "ethanoate" group. The ethyl group is a two-carbon chain, and the ethanoate group is a combination of a one-carbon chain and a carbonyl group (C=O) that is also attached to an oxygen atom. When the end alkyl groups are interchanged, the "ethyl" group is moved from the second carbon to the first carbon of the ethanoate group, and the "propanoate" group is formed. The "propanoate" group consists of a three-carbon chain and the carbonyl group. Therefore, the resulting compound is ethyl propanoate, which has a chemical formula of CH3CH2COOCH2CH3. This compound is commonly used as a flavoring agent and has a fruity odor reminiscent of pears.
Tambaya 29 Rahoto
The oxidation state(s) of nitrogen in ammonium nitrite is/are
Bayanin Amsa
Ammonium nitrite = NH4
NO2
NH+4
: Let the oxidation number of Nitrogen = x
x + 4 = 1 ⟹
x = 1 - 4
x = -3
NO−2
: x - 4 = -1
x = -1 + 4 ⟹
x = +3.
The oxidation numbers for Nitrogen in Ammonium Nitrite = -3, +3.
Tambaya 30 Rahoto
Consider the reaction
A(s) + 2B(g) → 2C(aq) + D(g)
What will be the effect of a decrease in pressure on the reaction?
Bayanin Amsa
Given: The equation below
A(s) + 2B(g) → 2C(aq) + D(g)
Since we have a higher number of moles of gaseous species on the LHS, i.e 2, a decrease in pressure will favor the forward reaction.
Tambaya 31 Rahoto
Hydrogen bond is a sort of
Bayanin Amsa
Hydrogen bond is a covalent intermolecular bond that exists between hydrogen and highly electronegative elements like nitrogen, oxygen and fluorine.
Tambaya 32 Rahoto
Which of the following is a physical change?
Bayanin Amsa
A physical change refers to a change in a substance that does not result in a change in its chemical composition. Out of the options provided, freezing ice cream is a physical change. This is because when ice cream is frozen, it changes from a liquid state to a solid state without any chemical reaction occurring. Exposing white phosphorus to air is a chemical change, as it reacts with oxygen in the air to form a new substance, phosphorus oxide. Burning kerosene is also a chemical change, as it undergoes combustion to form new substances, such as carbon dioxide and water vapor. Dissolving calcium in water is a physical change, as it simply involves the physical mixing of two substances without any chemical reaction occurring. Therefore, the only option that is a physical change is freezing ice cream.
Tambaya 34 Rahoto
Hydrocarbons which will react with Tollen's reagent conform to the general formula
Tambaya 35 Rahoto
Which of the following pairs cannot be represented with a chemical formula?
Bayanin Amsa
The pair that cannot be represented with a chemical formula is air and bronze. Air is a mixture of several gases, primarily nitrogen (N₂) and oxygen (O₂), with small amounts of other gases such as argon (Ar), carbon dioxide (CO₂), and neon (Ne). Since air is a mixture and not a pure substance, it cannot be represented by a chemical formula. Bronze, on the other hand, is an alloy composed mainly of copper (Cu) and tin (Sn) with small amounts of other metals. The composition of bronze can vary depending on the specific alloy, but it can be represented by a chemical formula such as CuSn. Sodium chloride (NaCl) is a compound composed of sodium (Na) and chlorine (Cl) in a fixed ratio of 1:1, and it can be represented by a chemical formula. Similarly, copper (Cu) and sodium chloride (NaCl) can each be represented by a chemical formula. Cu is an element, so its chemical formula is simply its symbol, while NaCl is a compound with a fixed ratio of sodium and chlorine atoms. Caustic soda (sodium hydroxide, NaOH) and washing soda (sodium carbonate, Na₂CO₃) are both compounds that can be represented by chemical formulas. NaOH consists of one sodium atom, one oxygen atom, and one hydrogen atom, while Na₂CO₃ consists of two sodium atoms, one carbon atom, and three oxygen atoms.
Tambaya 36 Rahoto
Which of the following represents the kind of bonding present in ammonium chloride?
Bayanin Amsa
Ammonium chloride contains both ionic and covalent bonds. In ammonium chloride, the ammonium ion (NH4+) is positively charged and the chloride ion (Cl-) is negatively charged. These ions are held together by ionic bonds, which are formed between positively and negatively charged ions. However, the bond between the hydrogen atom in the ammonium ion and the nitrogen atom in the ammonium ion is also a covalent bond. This type of covalent bond is known as a dative covalent bond, or a coordinate covalent bond, because the electron pair being shared is supplied by one atom only (the nitrogen atom in this case). So, the kind of bonding present in ammonium chloride is both ionic and dative covalent. In simple terms, ammonium chloride contains both ionic bonds between its positive and negative ions, and a dative covalent bond between the hydrogen atom and nitrogen atom within the ammonium ion.
Tambaya 37 Rahoto
Na2 CO3 + 2HCl → 2NaCl + H2 O + CO2
The indicator most suitable for this reaction should have a pH equal to
Bayanin Amsa
Methyl orange is the best indicator for the reaction with range 3.1 - 4.4.
Tambaya 38 Rahoto
Burning magnesium ribbon in air removes which of the following
(i) oxygen (ii) nitrogen (iii) argon and (iv) carbon(iv)oxide?
Bayanin Amsa
Burning magnesium ribbon in air will remove oxygen (option i) from the air, but not nitrogen (option ii), argon (option iii), or carbon dioxide (option iv). When magnesium burns, it reacts with oxygen in the air to form magnesium oxide. The reaction can be represented by the following equation: 2Mg(s) + O2(g) → 2MgO(s) The magnesium in the ribbon combines with oxygen in the air to form solid magnesium oxide. This reaction is exothermic, which means that it releases heat and light energy. So, when magnesium ribbon is burned in air, it consumes the oxygen in the air to form magnesium oxide. However, nitrogen, argon, and carbon dioxide are not chemically reactive with magnesium, and therefore are not removed from the air by the burning of magnesium ribbon. In summary, the correct option is (i) only - burning magnesium ribbon in air removes oxygen only.
Tambaya 39 Rahoto
When chlorine water is exposed to bright sunlight, the following products are formed
Tambaya 40 Rahoto
Which process(es) is/are involved in the turning of starch iodide paper blue-black by chlorine gas?
Bayanin Amsa
The process involved in the turning of starch iodide paper blue-black by chlorine gas is option number 3: chlorine oxidizes the iodide ion to produce iodine which attacks the starch to give the blue-black color. When chlorine gas comes in contact with iodide ions on the starch iodide paper, it oxidizes the iodide ion to form iodine. The iodine that is produced in this reaction is then able to react with the starch present on the paper to form a blue-black complex. This blue-black complex is formed due to the arrangement of the starch molecules and the iodine atoms in a way that causes them to absorb light at a specific wavelength, giving the blue-black color. Therefore, the blue-black color that is observed on the starch iodide paper is due to the reaction between iodine and starch, which is made possible by the oxidation of iodide ions by chlorine gas.
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